Since there are only three elements, you can certainly "just" check each case. Indeed, that's exactly what you should probably do: "Proof by cases" is a perfectly legitimate method of proof:
Consider each element as a separate "case" (there are only three to consider). For each element $x \in \{u, v, w\}$, find a counterexample which shows that $x$ cannot be the identity.
For example, suppose we test the case where $x = w$: From the table, we have that $w*w = v.\;$ This means $w$ cannot be an identity, since if $x = w$ were the identity, we must have $w \times w = w$. $\;\checkmark$
Do the same for each of $u$ and $v$, and you're done.
One observation, as noted below in the comments: See if you can prove that if an identity element $x$ exists, we would need to have one column for element $x$ replicate the left-most column, and the corresponding row for $x$ replicate the top most row, the "header" row.
Example where there exists an identity element $u$:
$$
\begin{array}{l}
\text{Example with identity u} \\
\begin{array}{c|ccc}
\hline
* & u & v & w \\
\hline
u & u & v & w \\
v & v & w & u \\
w & w & u & v \\
\hline
\end{array}
\end{array}
$$
Computer search finds these three completions and no others:
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & c & c & a \\
b & d & b & b & d \\
c & a & c & c & a \\
d & d & b & b & d \\
\end{array}
$$
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & a & c & c \\
b & b & b & d & d \\
c & c & c & a & a \\
d & d & d & b & b \\
\end{array}
$$
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & b & c & d \\
b & a & b & c & d \\
c & d & c & b & a \\
d & d & c & b & a \\
\end{array}
$$
Best Answer
$c$ is the identity and the operation is commutative so $$a\cdot c=c\cdot a=a$$ Similarly $$b\cdot c=c\cdot b=b$$ It seems to me that $a\cdot b$ can be $a$ or $b$.